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(A) The area $A$ of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is given by the integral:$A = \int_2^3 \sqrt{16-x^2} d

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$16 \sin^{-1}\left(\frac{3}{4}\right)$

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(A) The area $A$ of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is given by the integral:$A = \int_2^3 \sqrt{16-x^2} dx$Using the standard integral formula $\int \sqrt{a^2-x^2} dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}ight)$:$A = \left[\frac{x}{2}\sqrt{16-x^2} + 8\sin^{-1}\left(\frac{x}{4}ight)ight]_2^3$Evaluating the definite integral:$A = \left(\frac{3}{2}\sqrt{16-9} + 8\sin^{-1}\left(\frac{3}{4}ight)ight) - \left(\frac{2}{2}\sqrt{16-4} + 8\sin^{-1}\left(\frac{2}{4}ight)ight)$$A = \left(\frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}ight)ight) - \left(\sqrt{12} + 8\sin^{-1}\left(\frac{1}{2}ight)ight)$Since $\sqrt{12} = 2\sqrt{3}$ and $\sin^{-1}\left(\frac{1}{2}ight) = \frac{\pi}{6}$:$A = \frac{3}{2}\sqrt{7} + 8\sin^{-1}\left(\frac{3}{4}ight) - 2\sqrt{3} - 8\left(\frac{\pi}{6}ight)$$A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}ight)$Comparing this with the given expression $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+kight)$:Note that the given expression is $2 	imes \left(\frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + \frac{k}{2}ight)$.Actually,evaluating the integral result directly: $A = \frac{3}{2}\sqrt{7} - 2\sqrt{3} - \frac{4\pi}{3} + 8\sin^{-1}\left(\frac{3}{4}ight)$.To match the form $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+kight)$,we multiply the expression by $2$ and divide by $2$:$A = \frac{1}{2} \left(3\sqrt{7} - 4\sqrt{3} - \frac{8\pi}{3} + 16\sin^{-1}\left(\frac{3}{4}ight)ight)$Thus,$k = 16\sin^{-1}\left(\frac{3}{4}ight)$.

If the area of the region enclosed by the curve $x^2+y^2=16$ and the lines $x=2$ and $x=3$ is $\left(3 \sqrt{7}-4 \sqrt{3}-\frac{8 \pi}{3}+k\right)$ sq units,then $k$ equals

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